College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Physics Problem: Calculating the Magnitude of Final Velocity**
**Problem Statement:**
You threw a tennis ball from a balcony 4.1 meters above the ground upward at an angle of 45 degrees above horizontal with an initial speed of 4.0 meters per second. What was the magnitude of the total final velocity?
**Detailed Breakdown:**
1. **Initial Conditions:**
- Height of the balcony: 4.1 meters above the ground
- Launch angle: 45 degrees above the horizontal
- Initial speed: 4.0 meters per second
2. **Calculating the Initial Velocity Components:**
- Horizontal component (\(v_{0x}\)): \(v_{0x} = v_0 \cdot \cos(\theta)\)
\[
v_{0x} = 4.0 \, \text{m/s} \cdot \cos(45^\circ) = 4.0 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 2.83 \, \text{m/s}
\]
- Vertical component (\(v_{0y}\)): \(v_{0y} = v_0 \cdot \sin(\theta)\)
\[
v_{0y} = 4.0 \, \text{m/s} \cdot \sin(45^\circ) = 4.0 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 2.83 \, \text{m/s}
\]
3. **Vertical Motion Analysis:**
- Use the kinematic equation for vertical motion:
\[
y = v_{0y}t + \frac{1}{2} a t^2
\]
where \( y \) is the displacement (4.1 m), \( v_{0y} \) is the initial vertical velocity (2.83 m/s), \( a \) is the acceleration due to gravity (-9.8 m/s²), and \( t \) is the time of flight.
4. **Solving for Time of Flight:**
- Solving the quadratic equation:
\[
4.1 = 2.83t - \frac{1}{2}(9.8)t^](https://content.bartleby.com/qna-images/question/9e60ae64-4458-4b89-bb0e-7dd9dd5dc5fe/a95f53bb-112c-4b5d-aba6-100b8cf0133f/xjs5pnj_processed.png)
Transcribed Image Text:**Physics Problem: Calculating the Magnitude of Final Velocity**
**Problem Statement:**
You threw a tennis ball from a balcony 4.1 meters above the ground upward at an angle of 45 degrees above horizontal with an initial speed of 4.0 meters per second. What was the magnitude of the total final velocity?
**Detailed Breakdown:**
1. **Initial Conditions:**
- Height of the balcony: 4.1 meters above the ground
- Launch angle: 45 degrees above the horizontal
- Initial speed: 4.0 meters per second
2. **Calculating the Initial Velocity Components:**
- Horizontal component (\(v_{0x}\)): \(v_{0x} = v_0 \cdot \cos(\theta)\)
\[
v_{0x} = 4.0 \, \text{m/s} \cdot \cos(45^\circ) = 4.0 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 2.83 \, \text{m/s}
\]
- Vertical component (\(v_{0y}\)): \(v_{0y} = v_0 \cdot \sin(\theta)\)
\[
v_{0y} = 4.0 \, \text{m/s} \cdot \sin(45^\circ) = 4.0 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 2.83 \, \text{m/s}
\]
3. **Vertical Motion Analysis:**
- Use the kinematic equation for vertical motion:
\[
y = v_{0y}t + \frac{1}{2} a t^2
\]
where \( y \) is the displacement (4.1 m), \( v_{0y} \) is the initial vertical velocity (2.83 m/s), \( a \) is the acceleration due to gravity (-9.8 m/s²), and \( t \) is the time of flight.
4. **Solving for Time of Flight:**
- Solving the quadratic equation:
\[
4.1 = 2.83t - \frac{1}{2}(9.8)t^
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